 For rigid rotor we have discussed how one can set up Schrodinger equation starting from the square of angular momentum operator and we have worked out the phi dependent part and we have given you the answer for the theta dependent part and the answer we have discussed is wave function essentially is spherical harmonics theta phi is equal to the theta dependent part multiplied by phi dependent part and now we can talk about what is J and what is M this capital M is something that arises out of solution of your phi dependent part of the equation remember capital phi turned out to be a multiplied by e to the power i M phi what is a we will work out in the assignments what does M stand for it stands for the z component of angular momentum what are the allowed values 0 plus minus 1 plus minus 2 and so on and so forth yeah what is J J is another quantum number that comes when we work out the theta dependent part also do not forget that J the theta dependent part and phi dependent part are related by this M square so it also gives us a limit to the value of capital M okay so J turns out to be 0 1 2 3 4 0 and positive integers and the limit of capital M turns out to be J all right so if J equal to 3 for example you can have M equal to 0 plus minus 1 plus minus 2 plus minus 3 so we have 2 J plus 1 values of M okay we will come back to this once again when we talk about hydrogen atom the theta dependent part is basically a constant multiplied by a an associated legendary polynomial in cos theta okay that is what it is so the theta dependent part is a polynomial and the phi dependent part is an imaginary exponential term factor okay now what happens when we try to find the energy for that we go back to total angular momentum we are going to discuss angular momentum and its component in a little more detail in the next module for now just believe me when I write when the square of angular momentum operator operates on this spherical polar harmonics when the it operates on spherical harmonics I get back the same spherical harmonics wave function multiplied by its corresponding eigen value h cross square multiplied by J into J plus 1 so total angular momentum is h cross square multiplied by J into J plus 1 how did we obtain the Hamiltonian we obtain the Hamiltonian by dividing the square of angular momentum operator by 2 I so that 2 I is essentially a constant for the molecule so I can very simply write this Hamiltonian is L square by 2 mu r 0 square so this is the wave function and we know that this L square operates on the wave function to give us now I have written out h cross and I have written explicitly h by 2 pi square of that multiplied by J into J plus 1 okay this is a Schrodinger equation now I simply okay divided by your 2 I let us go back on that a little bit maybe this is a square from there we go to Hamiltonian and to get go to the Hamiltonian I have simply divided by what we had in the denominator we have 2 mu r 0 square so that gives us h square by 8 pi square mu 0 square so constant multiplied by J into J plus 1 multiplied by the wave function that is a Schrodinger equation this is the eigen value of energy in joule where J is equal to 0 1 2 3 so on and so forth as I had said alright so what do we learn from here before going there this is something that I write essentially because I am a spectroscopist generally spectroscopists do not want to work with joule they prefer centimeter inverse especially when working about rotational spectroscopy so generally we convert this to epsilon j is equal to h by 8 pi square i c multiplied by J into J plus 1 which is simply multiplying by h c and that is in centimeter inverse and it is simply written epsilon j equal to b into J into J plus 1 centimeter inverse where b is h by 8 pi square i c this is called the rotational constant so this is where we are alright let us take a look at this first of all v into J into J plus 1 what is the minimum value of J 0 what is the minimum value of energy then 0 so remember for a quantum harmonic oscillator energy could never be 0 because the oscillator if it is at rest then the position is 0 plus minus 0 x delta x delta px is also 0 plus minus 0 so delta x and delta px are both 0 and that violates uncertainty principle which is not allowed here what happens is let us say this is my rigid rotor okay this has a tail let us say this is my rigid rotor so it can rotate in any direction any theta phi can be span let us say it has come to rest okay it is true that the uncertainty in angular momentum is 0 what is the uncertainty in position for that we will need to know delta theta and delta phi well uncertainty theta is well theta can be anything right it can come to rest here or here or here anything anywhere so uncertainty in theta is effectively infinite of course it cannot be more than pi but in that domain that is infinity uncertainty in phi can be again anything between 0 and 2 pi yeah which is infinity so even though uncertainty in angular momentum is 0 uncertainty in the positional coordinates is infinity that is why the day is saved and a quantum rotor rigid rotor can come to rest anywhere in space okay uncertainty principle is not violated that is point number 1 so minimum rotational energy is 0 now look at this epsilon j equal to b into j into j plus 1 b equal to h by 8 pi square i c so what is j what is the next level after 0 j equal to 1 when you put j equal to 1 what do we get maybe I can just write that the minimum rotational energy is 0 we have written what happens when we will write like this these j values are already given here is not it what is the energy corresponding to j equal to 1 energy corresponding to j equal to 1 you can put j equal to 1 here so j equal to 1 j plus 1 equal to 2 so the energy turns out to be 2 b right 2 b what happens when j equal to 3 you can work that out j equal to 3 j plus 1 equal to 4 so sorry why did you go to j equal to 3 where it is okay 12 so 12 b I missed j equal to 2 here for some reason okay so you can work out the energies and the values are all written here now let us work out something interesting what is the energy gap between j equal to 1 and j equal to 0 obviously 2 b because this energy is 2 b this energy is 0 difference is 2 b what is the energy gap between j equal to 2 and j equal to 1 6 b minus 2 b so you can write it here the energy gap is 4 b what is the energy gap between j equal to 3 and j equal to 2 levels 12 b minus 6 b that is 6 b you see a pattern coming up j equal to 4 and j equal to 3 20 b minus 12 b that is 8 b and j equal to 5 and j equal to 4 30 b minus 20 b is 10 b okay not very difficult to work out really right because what I am trying to work out is delta epsilon for since I am a spectroscopist I always write in terms of spectroscopy j to j plus 1 transition that turns out to be equal to what you can work it out b into instead of j I will write j plus 2 so j plus 1 becomes well j plus 1 becomes j plus 2 and j becomes j plus 1 minus b j into j plus 1 so j plus 1 is common and inside the bracket you have j plus 2 minus j so it is 2 b into j plus 1 so the energy gaps turn out to be 2 b into j plus 1 okay so energy gap keeps increasing as you go higher up the ladder in rotational energy manifold does it remind you of something reminds me of particle in a box that is exactly what happened there also now this has profound implications in rotational spectroscopy turns out and I will not derive it here once again it is there in our molecular spectroscopy lectures one can work out the selection rule the selection rule turns out to be delta j equal to plus minus 1 which means that rotational lines in rotational spectra lines will occur at intervals of how much the first one will occur at 2 b corresponding to the delta g equal to plus minus 1 right so you can go from 0 to 1 that energy gap is 2 b then you cannot go from 0 to 2 but you can go from 1 to 2 remember delta is equal to plus minus 1 so that energy gap is 4 b then again you cannot go from 1 to 3 you can go from 2 to 3 that energy gap is 6 b the next one is 8 b next one is 10 b and so on and so forth so what you end up getting is spectra with lines that are equispaced for a rigid rotor so what we get is equispaced line spectra how is that useful it is useful because that difference in energy is 2 b into j plus 1 no sorry difference in energy is not 2 b into j plus 1 difference in energy is 2 b difference in lines difference in energy of lines is essentially 2 b so once you record a spectrum from the spacing you can work out b and if you work out b then you can work out this this is b so h is known pi is known c is known so knowing b you can work out i i remember is mu r 0 square again if you know which molecule you are working with mu is known so you can figure out what r 0 is from this spacing you can find out r 0 what is r 0 for a diatomic molecule it is a bound length so this rotational spectroscopy provides a means for a determination of bond length that is the application of the rigid rotor model that we have discussed so far of course is a simple model life is not so simple so it is very possible that the rotor is not rigid the molecule while rotating does not keep its bond length constant but all that comes into the domain of little higher level quantum mechanics and spectroscopy will not go into that right now it is discussed in our molecular spectroscopy course of course for now we close this discussion but it is not completely over because remember we still have to discuss angular momentum in a little more detail and in doing so we will learn some elegant features of quantum mechanics in the next module.